For a binomial variate X with parameters n=5 | vedclass

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(C) Given $n=5$ and $p=\frac{3}{4}$,so $q=1-p=\frac{1}{4}$.$\alpha = \frac{1}{9} P(X \geq 3) = \frac{1}{9} [P(X=3) + P(X=4) + P(X=5)]$$= \frac{1}{9} [

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(C) Given $n=5$ and $p=\frac{3}{4}$,so $q=1-p=\frac{1}{4}$.$\alpha = \frac{1}{9} P(X \geq 3) = \frac{1}{9} [P(X=3) + P(X=4) + P(X=5)]$$= \frac{1}{9} [^5C_3(\frac{3}{4})^3(\frac{1}{4})^2 + ^5C_4(\frac{3}{4})^4(\frac{1}{4})^1 + ^5C_5(\frac{3}{4})^5]$$= \frac{1}{9} [10 \cdot \frac{27}{1024} + 5 \cdot \frac{81}{1024} + 1 \cdot \frac{243}{1024}] = \frac{1}{9} [\frac{270+405+243}{1024}] = \frac{1}{9} [\frac{918}{1024}] = \frac{102}{1024}$.$\beta = P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$$= ^5C_0(\frac{3}{4})^0(\frac{1}{4})^5 + ^5C_1(\frac{3}{4})^1(\frac{1}{4})^4 + ^5C_2(\frac{3}{4})^2(\frac{1}{4})^3$$= 1 \cdot \frac{1}{1024} + 5 \cdot \frac{3}{1024} + 10 \cdot \frac{9}{1024} = \frac{1+15+90}{1024} = \frac{106}{1024}$.Now,$256(\beta-\alpha) = 256(\frac{106}{1024} - \frac{102}{1024}) = 256(\frac{4}{1024}) = 256(\frac{1}{256}) = 1$.

For a binomial variate $X$ with parameters $n=5$ and $p=\frac{3}{4}$,if $\alpha=\frac{1}{9} P(X \geq 3)$ and $\beta=P(X \leq 2)$,then $256(\beta-\alpha)=$

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